Question

$x^{{\log }_{5}x}>5$ implies:

• A. $x\in (0,\infty )$
• B. $x\in (0,\frac{1}{5})\cup (5,\infty )$
• C. $x\in (1,\infty )$
• D. $x\in (1,2)$

Answer 1

The correct option is B $x\in (0,\frac{1}{5})\cup (5,\infty )$

Taking logarithm with base $5$, we have

$lo{g}_5\left(x^{{\log }_{5}x}\right)>lo{g}_5\left(5\right)$

$\Rightarrow \left({\log }_{5}x\right)\left({\log }_{5}x\right)>1$

$\Rightarrow {\left({\log }_{5}x\right)}^2-1>0$

$\Rightarrow ({\log }_{5}x-1)({\log }_{5}x+1)>0$

$\Rightarrow \left({\log }_{5}x\right)>1$ OR $\Rightarrow \left({\log }_{5}x\right)<-1$

$\Rightarrow x>5^1,x<5^{-1}$

$\Rightarrow x>5,x<\frac{1}{5}$

Also we must have $x>0$.

Thus, $x\in (0,\frac{1}{5})\bigcup (5,\infty )$.